Practical right-triangle problems and interpreting diagrams | Year 9

01 / LEARN

Understand the idea.

Make the model visible

Turn the situation into a justified right-triangle model. A vertical object and level ground form perpendicular sides; a straight ladder, taut cable or line of sight may form the hypotenuse. Label what each side measures before choosing an equation.

Watch the starting level of a height measurement. A sight line from someone's eyes gives height above the eyes, not automatically height above the ground. Draw the eye-level horizontal and keep the eye height separate until the end.

Return to the real situation

Return the calculated length to its physical meaning, with units and sensible precision. A ladder's vertical reach must be less than its length. A ramp's sloping length must exceed its horizontal run. Such comparisons can expose a misplaced side label.

When an object moves, use a separate triangle for each position. Calculate the required quantities from their own measurements, retaining unrounded values, then compare them. A drop is the original height minus the final height.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A ladder

A 10 m ladder has its foot 6 m from a vertical wall on level ground. How high does it reach?

Worked solution

1
Identify the perpendicular legs and hypotenuse.
$h^{2} + 6^{2} = 10^{2}$
The vertical wall and level ground meet at a right angle; the ladder is the ten-metre hypotenuse.
2
Subtract the horizontal leg square.
$\begin{aligned} h^{2} & = 100 - 36 \\ = 64 \end{aligned}$
The unknown height is a leg, so its square is found by subtraction.
3
Take the positive root and interpret it.
$h = \sqrt{64} = 8 m$
The ladder reaches eight metres up the wall, less than its ten-metre length.

Answer

8 m

EXAMPLE 02

Include eye height

An observer’s eyes are 1.5 m above level ground, 18 m horizontally from a vertical tree. Its top is 32° above horizontal. Find the tree height to two decimal places.

Worked solution

1
Define the height above the observer's eyes.
$tan 32^{\circ} = \frac{h}{18}$
The eye-level horizontal is adjacent and the tree height above that level is opposite.
2
Calculate that vertical rise.
$\begin{aligned} h & = 18 tan 32^{\circ} \\ = 11.247648 \dots m \end{aligned}$
Tangent relates the vertical rise to the eighteen-metre horizontal distance.
3
Add the height of the observer's eyes.
$H = 1.5 + 18 tan 32^{\circ}$
The tree starts at ground level, one and a half metres below the eye-level horizontal.
4
Round the total tree height.
$\begin{aligned} H & = 12.747648 \dots m \\ \approx 12.75 m \end{aligned}$
Rounding occurs after adding the eye height, and the result reports height above ground.

Answer

12.75 m

03 / YOUR TURN

Try it yourself.

Check your understanding

A ladder changes from reaching 12 m high to 5 m high. Find the drop.

Show answer & reasoning

7 m

The drop is the decrease in vertical height: $12 - 5 = 7 m$ . The final height of five metres is a separate quantity.

Keep practising

6 worksheet questions · 20 marks

Open worksheet Worked solutions

Read the questions online

A 13 m ladder has its foot 5 m from a wall. Find its vertical reach.
3 marks
The same 13 m ladder slides so its foot is now 12 m from the wall. How far does the top drop?
4 marks
A ramp makes $7^{\circ}$ with level ground and has horizontal run 4.8 m. Find its sloping length to two decimal places.
3 marks
A vertical pole casts a 4.5 m shadow on level ground. The Sun’s elevation is $36^{\circ}$ . Find the pole height to two decimal places.
3 marks
A taut 30 m kite string begins at ground level and makes $40^{\circ}$ to horizontal. Find the kite height to two decimal places.
3 marks
Eyes are 1.6 m above level ground, 30 m horizontally from a building. Its top is at $25^{\circ}$ above horizontal. Find total height to two decimal places.
4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A 13 m ladder has its foot 5 m from a wall. Find its vertical reach.
3 marks
$h = \sqrt{169 - 25} = 12$ m.
12 m
The same 13 m ladder slides so its foot is now 12 m from the wall. How far does the top drop?
4 marks
Original height $= \sqrt{169 - 25} = 12$ m.
New height $= \sqrt{169 - 144} = 5$ m.
Drop $= 12 - 5 = 7$ m.
7 m
A ramp makes $7^{\circ}$ with level ground and has horizontal run 4.8 m. Find its sloping length to two decimal places.
3 marks
$cos 7^{\circ} = \frac{4.8}{L}$ .
$L = \frac{4.8}{cos 7^{\circ}}$ .
4.84 m
A vertical pole casts a 4.5 m shadow on level ground. The Sun’s elevation is $36^{\circ}$ . Find the pole height to two decimal places.
3 marks
$h = 4.5 tan 36^{\circ}$ .
Assume a vertical pole and a straight ray of sunlight.
3.27 m
A taut 30 m kite string begins at ground level and makes $40^{\circ}$ to horizontal. Find the kite height to two decimal places.
3 marks
$h = 30 sin 40^{\circ}$ .
The string is the hypotenuse.
19.28 m
Eyes are 1.6 m above level ground, 30 m horizontally from a building. Its top is at $25^{\circ}$ above horizontal. Find total height to two decimal places.
4 marks
Height above eyes $= 30 tan 25^{\circ}$ .
Total height $= 1.6 + 30 tan 25^{\circ}$ .
15.59 m

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Mathematics

Year 9 similarity, Pythagoras and right-triangle trigonometric ratios, with prerequisite revision. Non-right-triangle sine/cosine rules are not required.

Official curriculum source ↗

In this lesson

Understand the idea.

Make the model visible

Return to the real situation

See the method in action.

A ladder

Identify the perpendicular legs and hypotenuse.

Subtract the horizontal leg square.

Take the positive root and interpret it.

Include eye height

Define the height above the observer's eyes.

Calculate that vertical rise.

Add the height of the observer's eyes.

Round the total tree height.

Try it yourself.

Check your understanding

Keep practising